When $K M n O_{4}$ acts as an oxidizing agent and ultimately forms $M n O_{4}^{- 2}$ , $M n O_{2}$ , $M n_{2} O_{3}$ and $M n^{+ 2}$ , then the number of electrons transferred ( per Mn atom) in each case respectively is:

4, 3, 1, 5

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1, 5, 3, 7

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1, 3, 4, 5

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3, 5, 7, 1

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Solution

The correct option is C
1, 3, 4, 5

Check the oxidation states of Mn.
$⟹$ No of electrons transferred will be change in oxidation state

$+ 7 M n O_{4}^{-} + e^{-} ⟶ + 6 M n O_{4}^{- 2}$

$+ 7 M n O_{4}^{-} + 3 e^{-} ⟶ + 4 M n O_{2}$

$+ 7 M n O_{4}^{-} + 4 e^{-} ⟶ + 3 M n_{2} O_{3}$ ;

$+ 7 M n O_{4}^{-} + 5 e^{-} ⟶ + 2 M n^{2 +}$

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When KMnO4 acts as an oxidizing agent and ultimately formsMnO−24 ,MnO2 ,Mn2O3 andMn+2 , then the number of electrons transferred ( per Mn atom) in each case respectively is:

The correct option is C 1, 3, 4, 5 Check the oxidation states of Mn. ⟹ No of electrons transferred will be change in oxidation state +7MnO−4+e−⟶+6MnO−24 +7MnO−4+3e−⟶+4MnO2 +7MnO−4+4e−⟶+3Mn2O3; +7MnO−4+5e−⟶+2Mn2+

When $K M n O_{4}$ acts as an oxidizing agent and ultimately forms $M n O_{4}^{- 2}$ , $M n O_{2}$ , $M n_{2} O_{3}$ and $M n^{+ 2}$ , then the number of electrons transferred ( per Mn atom) in each case respectively is: